Why is voltage the same in a parallel circuit?

I know this is a very very basic question, but the story behind this idea seems to confuse me.

From the definition of volatage, voltage = work you have to do to move a unit charge through the resistor.

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When the circuit combine in parallel(say battery is 5V, R1=R2), the voltage drop on both resistor,

therefore "work" must occur on both resistor.

Now this is where confused come, since there is "work" on both resistor, (5V each),

combine these two work together we get 10V which is impossible because it is more than the source, but from my idea of work, this seems paradox to me.

Help me clarify this please.

Sorry for my English.

I think you are confusing voltage with power/energy ("work"?).

Putting one resistor across the voltage source will cause power to be dissipated in the resistor. Add a second resistor and power will be dissipated in the second resistor.

The voltage will not change. The current drawn from the voltage source will increase when there are two resistors.

Power = voltage x current
Current = voltage / resistance

Keith

I didn't confuse voltage with power.

what I confuse is since the voltage drop on every resistor why does it not equal to the total voltage from source, this idea is true when the circuit is in series.

From wikipedia

Voltage is equal to the work which would have to be done, per unit charge, against a static electric field to move the charge between two points.

A voltage may represent either a source of energy (electromotive force), or it may represent lost, used, or stored energy (potential drop).

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spark360z said:

I didn't confuse voltage with power.

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I think you are doing because the very first thing you said is "voltage = work". This is simply not true. "work" is synonymous with power, not voltage. A perfect voltage source maintains its voltage no matter what the load is. What will change is the current and hence the power (as power = voltage * current).

Keith

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I don't think the Wikepedia definition is helpful for explaining real circuits but it is not saying "voltage = work", but "voltage = work per unit charge ..."

spark360z said:

I didn't confuse voltage with power.

what I confuse is since the voltage drop on every resistor why does it not equal to the total voltage from source, this idea is true when the circuit is in series.

From wikipedia

Click to expand...

keith is right about everything, but let me try to explain it from little physical point of view:

- wikipedia is right, but you have to read carefully. That definition is saying about work in STATIC electric field, whereas in case of a parallel circuit we have current field. That are two different types of fields.

- General voltage definition defines voltage as a potential difference between TWO POINTS - These are points A and B in your circuit. So the work wikipedia refers to is the work that is required to move the unit charge between points A and B, and as you can see in your circuit, there are two paths the unit charge can be moved through - electric current arise in both paths!

-work is defined as force F that acts on certain path l, W = F*l,
force F = E*Q, where E is electric field intensity, and Q is electric charge,
it means W = E*Q*l , when you write the units you will see: J = V/m *C*m, meters cancel each other out, you will end up with J = V*C. It means 1V is potential difference (energy bareer) that requires 1J of energy for each 1C of charge to be moved across.